# The closure of a face of a convex set is a face of the closure of the set

Let $$C$$ be a convex set in a normed space and let $$F$$ be a face of $$C.$$ By a "face", it is understood a set such that if the segment $$u + (0,1)(v-u)$$ with two end points in $$C$$ intersects $$F$$ then $$u,v \in F$$ (it can be shown, in fact, that $$[u, v] \subset F$$). Note, a face is not required to be convex as is often done.

I know that faces are closed in the induced topology ($$F$$ is a closed subset of $$C$$) and I know that closures preserve convexity ($$\bar C$$ is both closed and convex). Naturally, I wonder if $$\bar F$$ is a face of $$\bar C.$$

Prove or provide a counterexample: $$\bar F$$ a face of $$\bar C.$$

The reciprocal conclusion is easy: if $$G$$ is a face of $$\bar C$$ then $$G \cap C$$ is easily shown to be a face of $$C$$ (by hypothesis, the defining property of face is satisfied for $$u,v \in \bar C,$$ so it is also satisfied for $$u,v \in C,$$ from which it follows that $$G \cap C$$ is a face of $$C$$).

Also, since faces are closed in the induced topology, it is also true that $$F = \bar F \cap C.$$

Finally, I'd be very happy if the result were true in general and will be happy if the result is true for convex faces (i.e. assuming $$F$$ is also convex).

Here is a counter-example: take $$C = (-1,+1)^2 \cup \{ (0,1) \}$$ with face $$F$$ given by $$F=\{(0,1)\}.$$ Then $$F$$ is closed and convex, but not a face of $$\bar C$$.

• Of course, taking a closed portion of the boundary that is not a face will do the trick, this seems to be a general way to create counter-examples. Much easier than I thought. Thanks. Commented Aug 19, 2022 at 15:19